Escape velocity
The escape velocity of a planet or indeed any other
gravitational system is the velocity that it would have to be given to escape from the
gravitational field of that planet. It could be a provided in a short burst of acceleration to
gain enough kinetic energy and would be unpowered after that.
If we consider a space
probe of mass m at the surface of a planet radius R then :
Gravitational potential
energy = -GMm/R the energy required to escape from the field is therefore
+GMm/R.
Kinetic energy to be applied = ½ mv
2 and so ½
(mv
2) = GMm/R. Therefore :
Escape velocity (ve) = (2GM/R)1/2 = (2Rgo)1/2
It is interesting to think about the maximum velocity
that might be reached by a meteorite falling onto the Earth's surface. If the meteorite starts
from infinity with zero velocity and is accelerated towards the Earth by the Earth's field
alone then you should be able to see that the greatest speed that it could have when it
reached the Earth is the escape velocity of the Earth.
Example problems
1. Calculate the escape velocity of the Earth given that go = 9.8ms-2 and that the radius of the Earth is 6.4x106 m
Escape velocity ve = (2Rgo) = (2x6.4x106x9.8) = 1.12x104 ms-1 = 11.2 kms-1
2. Calculate the radius of a planet with the an escape velocity of 6.2 ms-1 - about the speed of take off of a top class high jumper on Earth. Assume the planet has the same average density as the Earth - 5500 kgm-3.
Escape velocity ve = (2GM/R) therefore R = 2GM/ve2 =2G(4/3)pR3r/ve2
and so R = (ve2/2G(4/3)pr)1/2
Planet radius = (6.22/2x6.67x10-11x1.33xpx5500) = 3.53x103 m = 3.53 km!
The escape velocity is
also important if we want to find whether a planet can retain its atmosphere. The higher
velocity molecules will escape the gravitational pull if their velocity is greater than the
escape velocity of the planet.
Change in g with height
This is minimal until satellite orbit heights are reached. In fact for a height of some 200
km above the Earth's surface the g value has decreased from about 9.8 ms
-2
to around 7.5 ms
-2.
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